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3 years ago

Help me please and thank you

Physics

1 answer:

katovenus [111]3 years ago

3 0

Answer:

go to : www.planetresourses.com/test2.00/answers, ant type in that test name

Explanation:

yee

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How is sound produced? Name its three<br>characteristics.

nydimaria [60]

Answer:

1.loudness

2.pitch

3.shrillness

5 0

3 years ago

What is the action force and reaction force of two bumper cars collide

raketka [301]

<span>action is the one car hitting the other, reaction is the other car being pushed away</span>

7 0

3 years ago

HELP I NEED THIS ANSWERED AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

WITCHER [35]

Displacement is d

Vf² = Vi² + 2 g d

(-20²) = (+10²) + 2 (-9.8) d

-19.6 d = 300

d = -15.3 m

negative means lower

time is t

d = Vi t + 1/2 g t²

-15.3 = 10 t + (-4.9) t²

4.9 t² - 10 t -15.3 = 0

t = 3.06 s

Hope this helps -John

8 0

3 years ago

A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the

DiKsa [7]

Answer:

a). M = 20.392 kg

b). am = 0.56 $m/s^2$ (block), aM = 0.28 $m/s^2$ (bucket)

Explanation:

a). We got N = mg cos θ,

f = $\mu_s N$

= $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$ .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$ .....................(iv)

We got, N = mg cos θ

$f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$ ................(v)

Mg - 2T = M $a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

6 0

3 years ago

The idea is to get as much EMF produced from the sprinter running through it. If you were the Olympic coach on a year when there

never [62]

Answer:

the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

Explanation:

To determine the advice to the runners, let's use the Farad equation to and

fem = -N $\frac{ d \phi}{dt}$ = -N $\frac{ B A Cos \theta }{dt}$

how the runners are moving

fi = B l x

fem = -N B l v

therefore the advice we can give are:

* the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

3 0

3 years ago